RD Sharma Solutions for Class 12 Maths Chapter 17 Increasing and Decreasing Functions

RD Sharma books provide multiple-choice questions and very short answer types of questions, by which students can enhance their analytical thinking and time management skills. They have proven to be essential for learning the syllabus and building the confidence that is required to face their exams. Students are suggested to go through RD Sharma Solutions since these solutions are most repeated in the exams as well as in competitive exams.

Chapter 17 of RD Sharma Solutions for Class 12 Maths Increasing and Decreasing Functions explains monotonicity of functions. The exercise-wise solution of these topics are available here in PDF format, which can be easily downloaded by the students. These solutions are prepared by subject experts at BYJU’S in order to guide the students in their academics. Some of the essential topics of this chapter are listed below.

  • The solution of rational algebraic inequations
  • Strictly increasing functions
  • Strictly decreasing functions
  • Monotonic functions
  • Monotonically increasing function
  • Monotonically decreasing functions
  • Necessary and sufficient conditions for monotonicity
  • Finding the intervals in which a function is increasing or decreasing
  • Proving the monotonicity of a function on a given interval
  • Finding the interval in which a function is increasing or decreasing

RD Sharma Solutions Class 12 Maths Chapter 17 Increasing and Decreasing Functions:Download PDF Here

RD Sharma Class 12 Maths Solutions Chapter 17 Increasing And Decreasing Functions
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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 17 – Increasing and Decreasing Functions

Exercise 17.1 Page No: 17.10

1. Prove that the function f(x) = loge x is increasing on (0, ∞).

Solution:

Let x1, x2 ∈ (0, ∞)

We have, x1 < x2

⇒ loge x1 < loge x2

⇒ f (x1) < f (x2)

So, f(x) is increasing in (0, ∞)

2. Prove that the function f(x) = loga x is increasing on (0, ∞) if a > 1 and decreasing on (0, ∞), if 0 < a < 1.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 1

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 2

3. Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.

Solution:

Given,

f (x) = ax + b, a > 0

Let x1, x2 ∈ R and x1 > x2

⇒ ax1 > ax2 for some a > 0

⇒ ax1 + b> ax2 + b for some b

⇒ f (x1) > f(x2)

Hence, x1 > x2 ⇒ f(x1) > f(x2)

So, f(x) is increasing function of R

4. Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.

Solution:

Given,

f (x) = ax + b, a < 0

Let x1, x2 ∈ R and x1 > x2

⇒ ax1 < ax2 for some a > 0

⇒ ax1 + b < ax2 + b for some b

⇒ f (x1) < f(x2)

Hence, x1 > x2⇒ f(x1) < f(x2)

So, f(x) is decreasing function of R

Exercise 17.2 Page No: 17.33

1. Find the intervals in which the following functions are increasing or decreasing.

(i) f (x) = 10 – 6x – 2x2

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 3

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 4

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 5

(ii) f (x) = x2 + 2x – 5

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 6

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 7

(iii) f (x) = 6 – 9x – x2

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 8

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 9

(iv) f(x) = 2x3 – 12x2 + 18x + 15

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 10

(v) f (x) = 5 + 36x + 3x2 – 2x3

Solution:

Given f (x) = 5 + 36x + 3x2 – 2x3

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 11

⇒ f’(x) = 36 + 6x – 6x2

For f(x) now we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x2 = 0

⇒ 6(–x2 + x + 6) = 0

⇒ 6(–x2 + 3x – 2x + 6) = 0

⇒ –x2 + 3x – 2x + 6 = 0

⇒ x2 – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

Clearly, f’(x) > 0 if –2< x < 3 and f’(x) < 0 if x < –2 and x > 3

Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, –2) ∪ (3, ∞)

(vi) f (x) = 8 + 36x + 3x2 – 2x3

Solution:

Given f (x) = 8 + 36x + 3x2 – 2x3

Now differentiating with respect to x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 12

⇒ f’(x) = 36 + 6x – 6x2

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x2 = 0

⇒ 6(–x2 + x + 6) = 0

⇒ 6(–x2 + 3x – 2x + 6) = 0

⇒ –x2 + 3x – 2x + 6 = 0

⇒ x2 – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

Clearly, f’(x) > 0 if –2 < x < 3 and f’(x) < 0 if x < –2 and x > 3

Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, 2) ∪ (3, ∞)

(vii) f(x) = 5x3 – 15x2 – 120x + 3

Solution:

Given f(x) = 5x3 – 15x2 – 120x + 3

Now by differentiating above equation with respect x, we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 13

⇒ f’(x) = 15x2 – 30x – 120

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 15x2 – 30x – 120 = 0

⇒ 15(x2 – 2x – 8) = 0

⇒ 15(x2 – 4x + 2x – 8) = 0

⇒ x2 – 4x + 2x – 8 = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, – 2

Clearly, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0 if –2 < x < 4

Thus, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4)

(viii) f(x) = x3 – 6x2 – 36x + 2

Solution:

Given f (x) = x3 – 6x2 – 36x + 2

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 14

⇒ f’(x) = 3x2 – 12x – 36

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 3x2 – 12x – 36 = 0

⇒ 3(x2 – 4x – 12) = 0

⇒ 3(x2 – 6x + 2x – 12) = 0

⇒ x2 – 6x + 2x – 12 = 0

⇒ (x – 6) (x + 2) = 0

⇒ x = 6, – 2

Clearly, f’(x) > 0 if x < –2 and x > 6 and f’(x) < 0 if –2< x < 6

Thus, f(x) increases on (–∞,–2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (–2, 6)

(ix) f(x) = 2x3 – 15x2 + 36x + 1

Solution:

Given f (x) = 2x3 – 15x2 + 36x + 1

Now by differentiating above equation with respect x, we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 15

⇒ f’(x) = 6x2 – 30x + 36

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x2 – 30x + 36 = 0

⇒ 6 (x2 – 5x + 6) = 0

⇒ 3(x2 – 3x – 2x + 6) = 0

⇒ x2 – 3x – 2x + 6 = 0

⇒ (x – 3) (x – 2) = 0

⇒ x = 3, 2

Clearly, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0 if 2 < x < 3

Thus, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3)

(x) f (x) = 2x3 + 9x2 + 12x + 20

Solution:

Given f (x) = 2x3 + 9x2 + 12x + 20

Differentiating above equation we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 16

⇒ f’(x) = 6x2 + 18x + 12

For f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x2 + 18x + 12 = 0

⇒ 6(x2 + 3x + 2) = 0

⇒ 6(x2 + 2x + x + 2) = 0

⇒ x2 + 2x + x + 2 = 0

⇒ (x + 2) (x + 1) = 0

⇒ x = –1, –2

Clearly, f’(x) > 0 if –2 < x < –1 and f’(x) < 0 if x < –1 and x > –2

Thus, f(x) increases on x ∈ (–2,–1) and f(x) is decreasing on interval (–∞, –2) ∪ (–2, ∞)

2. Determine the values of x for which the function f(x) = x2 – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x2 – 6x + 9 where the normal is parallel to the line y = x + 5.

Solution:

Given f(x) = x2 – 6x + 9

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 17

⇒ f’(x) = 2x – 6

⇒ f’(x) = 2(x – 3)

For f(x) let us find critical point, we must have

⇒ f’(x) = 0

⇒ 2(x – 3) = 0

⇒ (x – 3) = 0

⇒ x = 3

Clearly, f’(x) > 0 if x > 3 and f’(x) < 0 if x < 3

Thus, f(x) increases on (3, ∞) and f(x) is decreasing on interval x ∈ (–∞, 3)

Now, let us find coordinates of point

Equation of curve is f(x) = x2 – 6x + 9

Slope of this curve is given by

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 18

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 18

3. Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 19

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 20

4. Show that f(x) = e2x is increasing on R.

Solution:

Given f (x) = e2x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 21

⇒ f’(x) = 2e2x

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 2e2x > 0

⇒ e2x > 0

Since, the value of e lies between 2 and 3

So, whatever be the power of e (that is x in domain R) will be greater than zero.

Thus f(x) is increasing on interval R

5. Show that f (x) = e1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 22

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 23

6. Show that f(x) = loga x, 0 < a < 1 is a decreasing function for all x > 0.

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 24

7. Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π).

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 25

8. Show that f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 26

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 27

9. Show that f(x) = x – sin x is increasing for all x ϵ R.

Solution:

Given f (x) = x – sin x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 28

⇒ f’(x) = 1 – cos x

Now, as given x ϵ R

⇒ –1 < cos x < 1

⇒ –1 > cos x > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

10. Show that f(x) = x3 – 15x2 + 75x – 50 is an increasing function for all x ϵ R.

Solution:

Given f(x) = x3 – 15x2 + 75x – 50

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 29

⇒ f’(x) = 3x2 – 30x + 75

⇒ f’(x) = 3(x2 – 10x + 25)

⇒ f’(x) = 3(x – 5)2

Now, as given x ϵ R

⇒ (x – 5)2 > 0

⇒ 3(x – 5)2 > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

11. Show that f(x) = cos2 x is a decreasing function on (0, π/2).

Solution:

Given f (x) = cos2 x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 30

⇒ f’(x) = 2 cos x (–sin x)

⇒ f’(x) = –2 sin (x) cos (x)

⇒ f’(x) = –sin2x

Now, as given x belongs to (0, π/2).

⇒ 2x ∈ (0, π)

⇒ Sin (2x)> 0

⇒ –Sin (2x) < 0

⇒ f’(x) < 0

Hence, condition for f(x) to be decreasing

Thus f(x) is decreasing on interval (0, π/2).

Hence proved

12. Show that f(x) = sin x is an increasing function on (–π/2, π/2).

Solution:

Given f (x) = sin x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 31

⇒ f’(x) = cos x

Now, as given x ∈ (–π/2, π/2).

That is 4th quadrant, where

⇒ Cos x> 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval (–π/2, π/2).

13. Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).

Solution:

Given f(x) = cos x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 32

⇒ f’(x) = –sin x

Taking different region from 0 to 2π

Let x ∈ (0, π).

⇒ Sin(x) > 0

⇒ –sin x < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in (0, π)

Let x ∈ (–π, o).

⇒ Sin (x) < 0

⇒ –sin x > 0

⇒ f’(x) > 0

Thus f(x) is increasing in (–π, 0).

Therefore, from above condition we find that

⇒ f (x) is decreasing in (0, π) and increasing in (–π, 0).

Hence, condition for f(x) neither increasing nor decreasing in (–π, π)

14. Show that f(x) = tan x is an increasing function on (–π/2, π/2).

Solution:

Given f (x) = tan x

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 33

⇒ f’(x) = sec2x

Now, as given

x ∈ (–π/2, π/2).

That is 4th quadrant, where

⇒ sec2x > 0

⇒ f’(x) > 0

Hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval (–π/2, π/2).

15. Show that f(x) = tan–1 (sin x + cos x) is a decreasing function on the interval (π/4, π /2).

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 34

16. Show that the function f (x) = sin (2x + π/4) is decreasing on (3π/8, 5π/8).

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 35

Thus f (x) is decreasing on the interval (3π/8, 5π/8).

17. Show that the function f(x) = cot–1 (sin x + cos x) is decreasing on (0, π/4) and increasing on (π/4, π/2).

Solution:

Given f(x) = cot–1 (sin x + cos x)

RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 36

18. Show that f(x) = (x – 1) ex + 1 is an increasing function for all x > 0.

Solution:

Given f (x) = (x – 1) ex + 1

Now differentiating the given equation with respect to x, we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 37

⇒ f’(x) = ex + (x – 1) ex

⇒ f’(x) = ex(1+ x – 1)

⇒ f’(x) = x ex

As given x > 0

⇒ ex > 0

⇒ x ex > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x > 0

19. Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).

Solution:

Given f(x) = x2 – x + 1

Now by differentiating the given equation with respect to x, we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 38

⇒ f’(x) = 2x – 1

Taking different region from (0, 1)

Let x ∈ (0, ½)

⇒ 2x – 1 < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in (0, ½)

Let x ∈ (½, 1)

⇒ 2x – 1 > 0

⇒ f’(x) > 0

Thus f(x) is increasing in (½, 1)

Therefore, from above condition we find that

⇒ f (x) is decreasing in (0, ½)  and increasing in (½, 1)

Hence, condition for f(x) neither increasing nor decreasing in (0, 1)

20. Show that f(x) = x9 + 4x7 + 11 is an increasing function for all x ϵ R.

Solution:

Given f (x) = x9 + 4x7 + 11

Now by differentiating above equation with respect to x, we get

⇒ 
RD Sharma Solutions for Class 12 Maths Chapter 17 Increaing and Decreasing Functions Image 39

⇒ f’(x) = 9x8 + 28x6

⇒ f’(x) = x6(9x2 + 28)

As given x ϵ R

⇒ x6 > 0 and 9x2 + 28 > 0

⇒ x6 (9x2 + 28) > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

Also, access RD Sharma Solutions for Class 12 Maths Chapter 17 Increasing and Decreasing Functions

Exercise 17.1 Solutions

Exercise 17.2 Solutions

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